Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author HossamDoma
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, InputReader in, PrintWriter out) {

            int n = in.nextInt();
            int m = in.nextInt();

            int[][] mat = new int[n][m];

            int[] row = new int[n], col = new int[m];

            Arrays.fill(row, Integer.MAX_VALUE);
            Arrays.fill(col, Integer.MAX_VALUE);

            for (int i = 0; i < n; ++i) {
                for (int j = 0; j < m; ++j) {
                    mat[i][j] = in.nextInt();
                    row[i] = Math.min(row[i], mat[i][j]);
                    col[j] = Math.min(col[j], mat[i][j]);
                }
            }

            //validity check

            boolean ok = true;

            for (int i = 1; i < n; ++i) {
                for (int j = 1; j < m && ok; ++j) {
                    ok = ok && ((mat[i][j] - mat[0][j]) == (mat[i][0] - mat[0][0]));
                }
            }

            if (!ok) {
                out.println(-1);
                out.flush();
                out.close();
                return;
            }


            int ans = 0;

            if (n <= m) {

                for (int i = 0; i < n; ++i) {
                    ans += row[i];
                }

                for (int j = 0; j < m; ++j) {
                    col[j] = mat[0][j] - row[0];
                    ans += col[j];
                }

            } else {

                for (int j = 0; j < m; ++j) {
                    ans += col[j];
                }

                for (int i = 0; i < n; ++i) {
                    row[i] = mat[i][0] - col[0];
                    ans += row[i];
                }

            }

            out.println(ans);

            StringBuilder sb = new StringBuilder();

            if (ans > 0) {

                for (int i = 0; i < n; ++i)
                    for (int x = 1; x <= row[i]; ++x)
                        sb.append("row " + (i + 1) + "\n");

                for (int j = 0; j < m; ++j)
                    for (int x = 1; x <= col[j]; ++x)
                        sb.append("col " + (j + 1) + "\n");

            }

            out.println(sb.toString().trim());
            out.flush();
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}